Analyse circular and geostationary orbits
Relate gravitational attraction to centripetal acceleration and identify the defining properties of a geostationary orbit.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 13. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use Newton's laws, circular motion, vectors, scientific notation and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What supplies the centripetal resultant for a satellite in a circular orbit?
What you will learn
- Analyse a circular orbit by equating gravity and the required centripetal resultant
- Relate orbital speed and radius to the central mass
- State the period, direction and equatorial plane of a geostationary orbit
Gravity supplies the orbital resultant
For a satellite of mass m orbiting a much larger mass M at radius r, the gravitational force is GMm/r². In a circular orbit, this supplies the inward centripetal resultant mv²/r.
Equating these expressions gives v² = GM/r. The satellite mass cancels: for a given central mass and orbital radius, the circular orbital speed does not depend on the satellite's mass. The orbit radius is measured from the centre of the planet.
Geostationary satellites remain above one longitude
A geostationary satellite has an orbital period of 24 hours, moves from west to east and travels in a circular orbit directly above Earth's Equator. Its angular speed matches Earth's rotation, so it remains above the same point on the surface.
A polar orbit or a period unlike Earth's rotation cannot remain fixed above one place. Do not confuse a satellite that merely passes over the same region with a geostationary satellite.
Pause and explain
Which description is required for a geostationary satellite?
Worked example
A satellite orbits a planet of mass 6.0 × 10²⁴ kg at radius 7.0 × 10⁶ m. Estimate its circular speed.
Show the worked solution
- Equate gravity to centripetal force to get v = √(GM/r).
- Substitute G = 6.67 × 10⁻¹¹ N m² kg⁻².
- Check that the result has units m s⁻¹.
Answer v ≈ √(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 7.0 × 10⁶) ≈ 7.6 × 10³ m s⁻¹.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Starting from gravity and centripetal force, show the equation for the speed of a satellite in a circular orbit.
Give me a hint
Set GMm/r² equal to mv²/r and cancel common factors carefully.
Compare my reasoning
- GMm/r² = mv²/r.
- Cancel m and multiply by r.
- v² = GM/r, so v = √(GM/r).
The circular orbital speed is v = √(GM/r).
Look for these in your work
- Two force expressions equated
- Satellite mass cancelled
- Speed relation derived
A planet's mass is unchanged and an orbit radius increases from r to 4r. Compare the circular orbital speeds.
Give me a hint
Use v² = GM/r and compare speeds, not squared speeds.
Compare my reasoning
- At 4r, v₂² = GM/(4r) = v₁²/4.
- Take the square root of the ratio.
- v₂ = v₁/2.
The circular orbital speed at 4r is half the speed at r.
Look for these in your work
- Radius ratio applied
- Square-root step included
- Speed ratio stated
A satellite is 24 hours from completing an orbit but travels west to east in a plane tilted away from the Equator. Is it geostationary? Explain.
Give me a hint
Check the complete definition, including orbital plane, rather than using period alone.
Compare my reasoning
- Its period matches 24 hours.
- A geostationary orbit must also be directly above the Equator and move west to east.
- The tilted orbital plane fails the equatorial condition, so it is not geostationary.
No. The 24-hour period and direction are insufficient because its orbit is not above the Equator.
Look for these in your work
- Period checked
- Equatorial requirement applied
- Conclusion justified
Common mistakes
- Using a satellite's height above a planet where orbital radius is needed
- Assuming orbital force depends on satellite mass after it cancels
- Describing any 24-hour orbit as geostationary without its direction and plane
What can you explain now?
What supplies the centripetal resultant for a satellite in a circular orbit?
Compare with the explanation
Gravitational attraction
The inward gravitational force bends the satellite's velocity into a circular path.
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Make it stick
How does equating gravity and centripetal force explain the speed of a circular orbit?
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